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If two random variables are uniformly distributed over a region, how do you in general find the joint PDF of those random variables?

For example, if $(X,Y)$ is distributed uniformly over the region $-2\leq x\leq 2$, $0\leq y\leq 1-x^2$, how could you derive the joint density function of $X$ and $Y$?

I believe you have to integrate, but with what integrand? Thank you.

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the easiest way is to calculate the Area of the domain region and, the joint pdf is its reciprocal

In the example you posted, the domain area is the following

$$\int_{-1}^{1}[1-x^2]dx=4/3$$

thus

$$f_{XY}(x,y)=\frac{3}{4}\mathbb{1}_{[-1;1]}(x)\cdot\mathbb{1}_{[0;1-x^2]}(y)$$

you can set also $x \in \mathbb{R}$ but thea area does not changes due to the fact that $y \ge 0$

enter image description here

why is the joint pdf the reciprocal of the area?

observing the drawing of the joint domain, you get that

$$C\int_{-1}^{1}\left[ \int_0^{1-x^2}dy \right]dx=1$$

that is $C=3/4$

Now observe that the above double integral is, geometrically, the volume of a solid figure with base the green region and constant height C that is your uniform density.

thus

$$V=1=C\cdot A$$

that means

$$C=\frac{1}{A}$$

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  • $\begingroup$ Thanks for the answer, but why is the joint pdf the reciprocal of the area? $\endgroup$
    – fmtcs
    Commented Oct 26, 2021 at 12:51
  • $\begingroup$ @fmtcs : the uniform density is constant over all the bivariate domain... It's a known fact that $$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(x,y)dxdy=C\cdot\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}dxdy=1$$ This integral can be viewed as the volume of the solid figure having the domain as a base and a constant density as height...thus the density is the reciprocal of the base area $\endgroup$
    – tommik
    Commented Oct 26, 2021 at 13:00
  • $\begingroup$ @fmtcs : I did some edits...your original question confused me... :( $\endgroup$
    – tommik
    Commented Oct 26, 2021 at 13:14

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